A) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\]
B) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}\]
C) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{3}}\]
D) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}\,}}\,-C{{H}_{2}}-C{{H}_{3}}\]
Correct Answer: A
Solution :
: The branched chain isomers of an alkane have lower boiling points than the straight chain isomer. This is because the branched chain isomers are more compact and tend to acquire nearly spherical shape. This decreases the effective surface area of contact with the adjacent molecules. As a result, the strength of the van der Waals forces in the branched chain alkanes are weaker than in the straight chain alkanes.
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